Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 
Explanation:
Elevation in boiling point:
where,
= boiling point of solution = ?
= boiling point of pure carbon disulfide= 
= boiling point constant =
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute = 35.0 g
= mass of solvent (carbon disulphide) = 397.7 g
= molar mass of solute = 70.0 g/mol
Now put all the given values in the above formula, we get:
Therefore, the change in boiling point is 
Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
Answer is: the ratio of the effusion rate is 1.59 : 1.
1) rate of effusion of carbon monoxide gas = 1/√M(CO).
rate of effusion of carbon monoxide gas = 1/√28.
rate of effusion of carbon monoxide gas = 0.189.
2) rate of effusion of chlorine = 1/√M(Cl₂).
rate of effusion of chlorine = 1/√70.9.
rate of effusion of chlorine = 0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine =
= 0.189 : 0.119 / ÷0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.
In carbohydrates the C:H:O is 1:2:1
